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I always see such signals (colored in the image) in a timing diagram. Now I've figured out that they're called:

  • low-activ and
    for me an OR-gate
  • high-active
    for me an AND-gate

signals.

But why isn't it all drawn as high-active signals (so that the blue area is visualized like the red one)!? What's the reason for that diversity ?

enter image description here

Jorgos
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    The signals are drawn as they would appear on an oscilloscope. It makes comparison with actual scope capture much easier. I don't know if your question has a definitive answer, but I like the signals the way they are. – user57037 Dec 12 '15 at 22:58
  • Ditto squared @mkeith - get used to it - this is EE and we like to see things as they are. – Andy aka Dec 13 '15 at 00:28

2 Answers2

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The graph directly represents the voltage that you apply to the pin. If it were flipped, it would be confusing because it would no longer be a graph of the voltage on the pin.

If you mean to say "why are some signals active low", that is a more complicated question and has many possible answers.

Sometimes it is more convenient to have a signal as active low, such as a reset signal that will rise as the power supply comes up. This might give you the opportunity to put a resistor and capacitor on the pin making a reset delay circuit.

Other times, it may be an "open collector" type signal that can have many devices tied to it, creating a NOR function, or even just allowing a very simple to implement shared signal.

A third possibility is a combination signal such as a read/write direction pin in a bus, where in one state it is read and the other state it is write. That is often represented as \$R/\overline{W}\$.

Daniel
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The reason low-active signals aren't visualized as high-active is that it has an actual physical meaning. The output of a system with one high active enable input, only gives an output if the signal is high. Of course, this is exactly the same for low active signals but then the other way around.

If you would then visualize low active as high active, you would mix it all up and your circuitry would probably fail.

rzonnie
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